Abeaker contains 47.0 ml of a 0.47 m hi solution and is titrated with 25.0 ml of a 0.25 m of koh. what is the final ph, and where is this point in the titration?

-

1.8

End point passed.

Explanation: -

Volume of HI solution = 47.0 mL = 0.047 L

Strength of HI solution = 0.47 M

Since HI is a strong acid, all of HI will dissociate to give H +.

[H+ ] =0.47 M x 0.047 L

= 0.02209 mol

Volume of KOH = 25.0 mL = 0.025 L

Strength of KOH = 0.25 M

Since KOH is a strong base, all of KOH will dissociate to give OH-.

[OH-] = 0.25 M x 0.025L

= 0.00625 mol

Since [H+] and [OH-] react to form water,

[H+] unreacted = 0.02209 – 0.00625 = 0.01584 mol

Using the formula

pH = - log [H+]

= - log 0.01584

= 1.8

As the strong acid HI is being titrated by strong base KOH, the pH at the end point should be 7.

The pH has already crossed that. Thus the titration end point has already passed

b. atoms in metallic solids are not arranged in a regular pattern

explanation: